35
19 May 2010 Turbulence Modeling Lecture # 05 Turbulence Modelling JetWings 1

Lecture # 05 Turbulence Modelling

Embed Size (px)

Citation preview

Page 1: Lecture # 05 Turbulence Modelling

19 May 2010

Turbulence Modeling

Lecture # 05 Turbulence Modelling

JetWings

1

Page 2: Lecture # 05 Turbulence Modelling

19 May 2010

Outline1. Characteristics of turbulence 2. Approaches to predicting turbulent flows 3. Reynolds averaging4. RANS equations and unknowns5. The Reynolds-Stress Equations6. The closure problem of turbulence7. Characteristics of wall-bound turbulent flows8. Turbulence models and ranges of applicability

7.1. RANS7.2. LES/DES7.3. DNS

9. Example: diffuser

Lecture # 05 Turbulence Modelling

JetWings

2

Page 3: Lecture # 05 Turbulence Modelling

19 May 2010

Characteristics of turbulence • Randomness and fluctuation:

• Nonlinearity: Reynolds stresses from the nonlinear convective terms

• Diffusion: enhanced diffusion of momentum, energy etc.

• Vorticity/eddies/energy cascade: vortex stretching

u U u′= +

Lecture # 05 Turbulence Modelling

JetWings

3

Page 4: Lecture # 05 Turbulence Modelling

19 May 2010

•Dissipation: occurs at smallest scales •Three-dimensional: fluctuations are always 3D •Coherent structures: responsible for a large part of the mixing •A broad range of length and time scales: makingDNS very difficult

Characteristics of turbulence

Lecture # 05 Turbulence Modelling

JetWings

4

Page 5: Lecture # 05 Turbulence Modelling

19 May 2010

Approaches to predicting turbulent flows• AFD, EFD and CFD:

– AFD: No analytical solutions exist – EFD: Expensive, time-consuming, and sometimes impossible

(e.g. fluctuating pressure within a flow) – CFD: Promising, the need for turbulence modeling

• Another classification scheme for the approaches– The use of correlations:– Integral equations: reduce PDE to ODE for simple cases – One-point closure: RANS equations + turbulent models – Two-point closure: rarely used, FFT of Two-point equations – LES: solve for large eddies while model small eddies – DNS: solve NS equations directly without any model

( )ReDC f=

Lecture # 05 Turbulence Modelling

JetWings

5

Page 6: Lecture # 05 Turbulence Modelling

19 May 2010

Deeper insights on RANS/URANS/LES

RANS

URANS

LES

Lecture # 05 Turbulence Modelling

JetWings

6

Page 7: Lecture # 05 Turbulence Modelling

19 May 2010

Reynolds Averaging

• Time averaging: for stationary turbulence

• Spatial averaging: for homogenous turbulence

• Ensemble averaging: for any turbulence

• Phase averaging: for turbulence with periodic motion

Lecture # 05 Turbulence Modelling

JetWings

7

Page 8: Lecture # 05 Turbulence Modelling

19 May 2010

• RANS equation

• RANS equation in conservative form

• Numbers of unknowns and equations–Unknowns: 10= P (1) + U (3) + (6) –Equations: 4 = Continuity (1) + Momentum (3)

RANS equations and unknowns

( )2i ij ji

jj i

i j

U U PU S

t xu u

x xρ ρ µ ρ∂ ∂ ∂ ∂+ = − +

∂ ∂ ∂′ ′−

( ) ( )2ij i ji

j i jj i

U PU U S

t xu

x xuρ ρ µ′ ′∂ ∂ ∂ ∂+ + = − +

∂ ∂ ∂ ∂

( )i ju u′ ′−

Lecture # 05 Turbulence Modelling

JetWings

8

Page 9: Lecture # 05 Turbulence Modelling

19 May 2010 9

The Reynolds-Stress Equation•Derivation: Taking moments of the NS equation.Multiply the NS equation by a fluctuating property and time average the product. Using this procedure, one can derive a differential equation for the Reynolds-stress tensor.

2ij ij j ik ik jk

k k k

ji

k k

uu

x x

U UU

t x x x

τ ττ τ ν

∂ ∂ ∂ ∂+ = − −′

+∂ ∂ ∂ ∂

∂′∂∂ ∂

jii j k

j

ij

ki k

uu p pu u u

x x x xν

ρ ρτ ∂ ∂+ + +

′′ ′ ′∂ ∂ ′ ′ ′∂ ∂

+∂ ∂

•NEW Equations: 6 = 6 equation for the Reynolds stress tensor •NEW Unknowns: 22= 6 + 6 + 10

2 6ji

k k

uuunkowns

x xν

′∂′∂ →∂ ∂

6ji

j i

uu p punkowns

x xρ ρ ′′ ′ ′∂ ∂+ → ∂ ∂

10i j ku u u unkowns′ ′ ′→

Lecture # 05 Turbulence Modelling

JetWings

9

Page 10: Lecture # 05 Turbulence Modelling

19 May 2010

The closure problem of turbulence• Because of the non-linearity of the Navier-

Stokes equation, as we take higher and higher moments, we generate additional unknowns at each level.

• In essence, Reynolds averagingis a brutal simplification that loses much of the informationcontained in the Navier-Stokes equation.

• The function of turbulence modelingis to devise approximations for the unknown correlationsin terms of flow properties that are known so that a sufficient number of equations exist.

• In making such approximations, we close the system.

Lecture # 05 Turbulence Modelling

JetWings

10

Page 11: Lecture # 05 Turbulence Modelling

19 May 2010

Turbulent models (RANS) • Boussinesq eddy-viscosity approximation• Algebraic (zero-equation) models

– Mixing length – Cebeci-Smith Model – Baldwin-Lomax Model

• One-equation models– Baldwin-Barth model – Spalart-Allmaras model

• Two-equation models – k-ε model – k-ω model

• Four-equation (v2f) models • Reynolds-stress (seven-equation) models

Lecture # 05 Turbulence Modelling

JetWings

11

Page 12: Lecture # 05 Turbulence Modelling

19 May 2010

Turbulent models (RANS)

2

3j

Ti

i j ijj i

UUu u

x xkδν

∂∂′ ′− = + − ∂ ∂

• Boussinesq eddy-viscosity approximation

• Dimensional analysis shows: , where q is a turbulence velocity scale and L is a turbulence length scale. Usually where is the turbulent kinetic energy. Models that do not provide a length scale are called incomplete.

T C qLµν =

2q k=1

2 i ik u u′ ′=

Lecture # 05 Turbulence Modelling

JetWings

12

Page 13: Lecture # 05 Turbulence Modelling

19 May 2010

Turbulent models (0-eqn RANS)

• Mixing length model: 2T mix

dUl

dyν =

• Assume for free shear flow, then( )mixl xαδ=

α=0.180 for far wake

α=0.071 for mixing layer

α=0.098 for plane jet

α=0.080 for round jet • Comments:

Reliable only for free shear flows with different α values

Not applicable to wall-bounded flows

Lecture # 05 Turbulence Modelling

JetWings

13

Page 14: Lecture # 05 Turbulence Modelling

19 May 2010

Turbulent models (0-eqn RANS) Cebeci-Smith Model (Two-layer model)

,i

o

T m

TT m

y y

y y

νν

ν≤= >

Where is the smallest value of for my yi oT Tν ν=

Closure coefficients:

Outer layer:

Inner layer:1 22 2

2

iT mix

U Vl

y xν

∂ ∂ = + ∂ ∂

01 y Amixl y eκ

+ +− = −

( )0

* ;T e v KlebU F yν α δ δ=

0.40κ = 0.0168α =1 2

226 1

dP dxA y

Uτρ

−+

= +

( )16

; 1 5.5Kleb

yF y δ

δ

− = +

( )*

01v eU U dy

δδ = −∫Comments:

Three key modifications to mixing length model; Applicable to wall-bounded flows;Applicable to 2D flows only;Not reliable for separated flows;

difficult to determine in some cases;*, ,v eUδ δ

Lecture # 05 Turbulence Modelling

JetWings

14

Page 15: Lecture # 05 Turbulence Modelling

19 May 2010

Turbulent models (0-eqn RANS) Baldwin-Lomax Model (Two-layer model)

,i

o

T m

TT m

y y

y y

νν

ν≤= >

Where is the smallest value of for my yi oT Tν ν=

Closure coefficients:

Outer layer:

Inner layer:

Comments:

Applicable to 3D flows;Not reliable for separated flows; No need to determine *, ,v eUδ δ

2

iT mixlν ω= 01 y Amixl y eκ

+ +− = −

( )0 max;T cp wake Kleb KlebC F F y y Cν α=

0.40κ = 0.0168α = 26A+ = 1.6cpC = 0.3KlebC = 1wkC =

( )16

maxmax

; 1 5.5Kleb KlebKleb

yF y y C

y C

− = +

1 22 2 2V U W V U W

x y y z z xω

∂ ∂ ∂ ∂ ∂ ∂ = − + − + − ∂ ∂ ∂ ∂ ∂ ∂

Lecture # 05 Turbulence Modelling

JetWings

15

Page 16: Lecture # 05 Turbulence Modelling

19 May 2010

Turbulent models (1-eqn RANS) Baldwin-Barth model

Kinematic eddy viscosity :

Turbulence Reynolds number :

Closure coefficients and auxiliary relations:

1 2T TC R D Dµν ν= %

( ) ( ) ( ) ( ) ( ) ( )2

2 2 1

1T T T

j Tk

TT

kT

k kj

R R RU C f C R P

t x x

R

x x xεε ε ε

νν ν νν ν ν σ ν

σ∂ ∂ ∂

+ = − + +∂

−∂

∂∂ ∂∂ ∂∂

% % %

%

%

1 1.2Cε = 2 2.0Cε = 0.09Cµ = 0 26A+ = 2 10A+ = ( )1 2 2

1 CC C µ

ε εεσ κ

= − 0.41κ =

2

3ji i k k

Tj i j k k

UU U U UP

x x x x xν

∂∂ ∂ ∂ ∂= + − ∂ ∂ ∂ ∂ ∂

01 1 y AD e

+ +−= − 22 1 y AD e

+ +−= −

0 21 1 2 12 1 2 1 2

2 2 0 21 2

1 11 y A y AC C D D

f D D D D e eC C y A AD D

ε ε

ε ε κ+ + + +− −

+ + +

= + − + ⋅ + +

Lecture # 05 Turbulence Modelling

JetWings

16

Page 17: Lecture # 05 Turbulence Modelling

19 May 2010

Turbulent models (1-eqn RANS) Spalart-Allmaras model

Kinematic eddy viscosity :

Eddy viscosity equation:

Closure coefficients and auxiliary relations:

1T vfν ν= %

( )2

11 1

1 bj b w w

j kk kk

cU c S c f

t x d xx x x

νν νσ

ν ν ν ν ννσ

∂ ∂+ ∂ ∂ ∂ ∂ + = − + + ∂ ∂ ∂ ∂∂ ∂

% % % % %% %

%%

1 0.1355bc = 2 0.622bc = 1 7.1vc = 2 3σ = ( )211 2

1 bbw

ccc

κ σ+

= +

2 0.3wc = 2 2.0wc = 0.41κ =

3

1 3 31

vv

fc

χχ

=+ 2

1

11v

v

ff

χχ

= −+

1 663

6 63

1 ww

w

cf g

g c

+= +

νχν

=% ( )6

2wg r c r r= + − 2 2r

S d

νκ

=%

%

22 2 vS S fd

νκ

= +%

% 2 ij ijS = Ω Ω

Lecture # 05 Turbulence Modelling

JetWings

17

Page 18: Lecture # 05 Turbulence Modelling

19 May 2010

Turbulent models (1-eqn RANS)

Comments on one-equation models:1.One-equation models based on turbulence kinetic energy are incompleteas

they relate the turbulence length scales to some typical flow dimension. They are rarely used.

2.One-equation models based on an equation for the eddy viscosity are completesuch as Baldwin-Barth model and Spalart-Allmaras model.

3.They circumvent the need to specify a dissipation length by expressing the decay, or dissipation, of the eddy viscosity in terms of spatial gradients.

4 Spalart-Allmaras model can predicts better results than Baldwin-Barth model, and much better results for separated flow than Baldwin-Barth model and algebraic models.

5 Also most of DES simulations are based on the Spalart-Allmaras model.

Lecture # 05 Turbulence Modelling

JetWings

18

Page 19: Lecture # 05 Turbulence Modelling

19 May 2010

Turbulent models (2-eqn RANS) k-ε model:

2T C kµν ε=

( )ij ij T k

j j j j

Uk k kU

t x x x xτ ε ν ν σ

∂∂ ∂ ∂ ∂+ = − + + ∂ ∂ ∂ ∂ ∂

( )2

1 2i

j ij Tj j j j

UU C C

t x k x k x xε ε εε ε ε ε ετ ν ν σ

∂∂ ∂ ∂ ∂+ = − + + ∂ ∂ ∂ ∂ ∂

1 1.44Cε =2 1.92Cε = 0.09Cµ = 1.0kσ = 1.3εσ =

,

,

,

,

k-ω model:( )C kµω ε= 3 2l C kµ ε=

T kν ω=

( )* *ij ij T

j j j j

Uk k kU k

t x x x xτ β ω ν σ ν

∂∂ ∂ ∂ ∂+ = − + + ∂ ∂ ∂ ∂ ∂

( )2ij ij T

j j j j

UU

t x k x x x

ω ω ω ωα τ βω ν σν ∂∂ ∂ ∂ ∂+ = − + + ∂ ∂ ∂ ∂ ∂

13

25α =

0 fββ β= ** *

0 fβ

β β= 1

2σ = * 1

2σ = 0

9

125β = 1 70

1 80f ω

βω

χχ

+=+

( )3*0

ij jk kiSωχ

β ω

Ω Ω= *

0

9

100β = *

2

2

1, 0

1 6800

1 400

k

kk

k

χχ χχ

≤= + > +

3

1k

j j

k

x x

ωχω

∂ ∂=∂ ∂

* kε β ω= 1 2l k ω=

Lecture # 05 Turbulence Modelling

JetWings

19

Page 20: Lecture # 05 Turbulence Modelling

19 May 2010

Turbulent models (2-eqn RANS)

,

Comments on two-equation models:1. Two-equation models are complete;2. k-ε and k-ω models are the most widely used two-

equation models and a lot of versions exist. For example, a popular variant of k-ω model introduced by Menter has been used in our research code CFDSHIP-IOWA. There are also a lot of low-Reynolds-number versions with different damping functions.

3. k-ω model shows better results than k-ε model for flows with adverse pressure gradient and separated flows as well as better numerical stability.

Lecture # 05 Turbulence Modelling

JetWings

20

Page 21: Lecture # 05 Turbulence Modelling

19 May 2010

Turbulent models (4-eqn RANS) v2f-kε model:

,

,

,

,

v2f-kω model:

2T C v Tµν =

( )2 2 2

tj j

Dv v vkf

Dt k x xε ν ν

∂ ∂= − + + ∂ ∂

22 2 1

2

2

3kPC v

L f f CT k k

∇ − = − −

t

j k j

Dk kP

Dt x x

νε νσ

∂ ∂= − + + ∂ ∂

2

1 2t

j j

DC P C

Dt k k x xε εε

νε ε ε ενσ

∂ ∂= − + + ∂ ∂

1 0.4C = 2 0.3C = 0.3LC = 70Cη = 2 1.9Cε = 1.0εσ =( )

42

1 1.3 0.25 1 2LC C d Lε = + + max ,6

kT

νε ε =

1 43 2 3

max ,L

kL C Cη

νε ε

=

12

nn

T C k vµν ω−

=

( )2 2

26 tj j

Dv vkf v

Dt k x x

ε ν ν ∂ ∂= − + +

∂ ∂ ( )

2 22 2

1 2

1 2 11 5

3kPv v

L f C CT k T k k

′∇ − = − − − −

*i tij

j j k j

UDk kk

Dt x x x

ντ β ω νσ

∂ ∂ ∂= − + + ∂ ∂ ∂

12

2

n

i tij

j j j

UD v

Dt k x k x xω

νω ω ωα τ βω νσ

− ∂ ∂ ∂= − + + ∂ ∂ ∂

0.45977α = * 0.09β = 3 40β = 1.0kσ = 1.5ωσ =

max ,6k

ε ε =

1 43 2 3

max ,L

kL C Cη

νε ε

=

Lecture # 05 Turbulence Modelling

JetWings

21

Page 22: Lecture # 05 Turbulence Modelling

19 May 2010

Turbulent models (4-eqn RANS)

,

Comments on four-equation models:1. For two-equation models a major problem is that it is hard to

specify the proper conditions to be applied near walls. 2. Durbin suggested that the problem is that the Reynolds number is

low near a wall and that the impermeability condition (zero normal velocity) is far more important. That is the motivation for the equation for the normal velocity fluctuation.

3. It was found that the model also required a damping function f, hence the name v2f model.

4. They appear to give improved results at essentially the same cost

as the k-ε and k-ω models especially for separated flows. Hopefully the

v2f-kω model can have better numerical stability than v2f-kε model as their counterparts behave in two-equation models.

Lecture # 05 Turbulence Modelling

JetWings

22

Page 23: Lecture # 05 Turbulence Modelling

19 May 2010

Turbulent models (7-eqn RANS) Some of the most noteworthy types of applications for which models based on the Boussinesq approximation fail are:

1. Flows with sudden changes in mean strain rate2. Flow over curved surfaces3. Flow in ducts with secondary motions4. Flow in rotating fluids5. Three-dimensional flows6. Flows with boundary-layer separation

In Reynolds-stress models, the equations for the Reynolds stress

tensor are modeled and solved along with the ε-equation:

2ij ij j ijik ik jk

k

j ji ii j k

k k jk kik k

u uu u p pu

U UU

t x xu

xu

x x x x x x

τ τ ττ ν

ρ ρτ ν

′ ′∂′ ′ ′ ′∂ ∂ ∂ ′ ′ ′+∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂+ = − − + + + + ∂ ∂ ∂ ∂ ∂ ∂

Lecture # 05 Turbulence Modelling

JetWings

23

Page 24: Lecture # 05 Turbulence Modelling

19 May 2010

Turbulent models (7-eqn RANS) These are some versions of Reynolds stress models:

1. LRR rapid pressure-strain model 2. Lumley pressure-strain model3. SSG pressure-strain model 4. Wilcox stress- model

Comments on Reynolds stress models:1.Reynolds stress models require the solution of seven additional

PDEsand those equation are even harder to solve than the two-equation models.

2. Although Reynolds stress models have greater potential to represent turbulent flow more correctly, their success so far has been moderate.

3. There is a lot of current research in this field, and new models are often proposed. Which model is best for which kind of flow is not clear due to the fact that in many attempts to answer this question numerical errors were too large to allow clear conclusions to be reached.

Lecture # 05 Turbulence Modelling

JetWings

24

Page 25: Lecture # 05 Turbulence Modelling

19 May 2010

Turbulent models (LES)

• Large scale motions are generally much more energetic than the small scale ones.

• The size and strength of large scale motions make them to be the most effective transporters of the conserved properties.

• LES treats the large eddies more exactly than the small ones may make sense

• LES is 3D, time dependent and expensive but much less costly than DNS.

Lecture # 05 Turbulence Modelling

JetWings

25

Page 26: Lecture # 05 Turbulence Modelling

19 May 2010

Turbulent models (LES, Scale-similarity model and Dynamic model)

• Dynamic model: 1. filtered LES solutions can be filtered again using a filter broader than the previous filter to obtain a very large scale field.2. An effective subgrid-scale field can be obtained by subtraction of the two fields.3. model parameter can then be computed.Advantages: 1. model parameter computed at every spatial grid point and every time step from the results of LES2. Self-consistent subgrid-scale model3. Automatically change the parameter near the wall and in shear flowsDisadvantages: backscatter (eddy viscosity<0) may cause instability.

Lecture # 05 Turbulence Modelling

JetWings

26

Page 27: Lecture # 05 Turbulence Modelling

19 May 2010

Turbulent models (DES)

• Massively separated flows at high Re usually involveboth large and small scale vortical structures and verythin turbulent boundary layer near the wall

• RANS approaches are efficient inside the boundarylayer but predict very excessive diffusion in the separatedregions

• LES is accurate in the separated regions but is unaffordablefor resolving thin near-wall turbulent boundary layers atindustrial Reynolds numbers

• Motivation for DES: combination of LES and RANS. RANSinside attached boundary layer and LES in the separatedregions

Lecture # 05 Turbulence Modelling

JetWings

27

Page 28: Lecture # 05 Turbulence Modelling

19 May 2010

DES Formulation

* 3 2

3 2

/kRANS k

kDES

D k k l

kDl

ωρβ ω ρ

ρ−= =

=%

• Modification to RANS models was straightforward by substituting the length scale , which is the distance to the closest wall, with the new DES length scale, defined as:

• where is the DES constant, is the grid spacing and is based on the largest dimensions of the local grid cell, and are the grid spacing in x, y and z coordinates respectively• Applying the above modification will result in S-A Based, standard k-ε (or k-ω) based and Menter’s SST based DES models, etc.

l%

min( , )w DESl d C= ∆% max( , , )x y zδ δ δ∆ =

wd

DESC ∆, ,x y zδ δ δ

1 2 *( )

min( , )

k

k DES

l k

l l C

ω

ω

β ω−

=

= ∆%

Lecture # 05 Turbulence Modelling

JetWings

28

Page 29: Lecture # 05 Turbulence Modelling

19 May 2010

Resolved/Modeled/Total Reynolds stress (DES)

TKE

Resolved

Modeled

Total

Modeled Reynolds stress:

∂∂+

∂∂−=

x

v

y

uvu tν''

Lecture # 05 Turbulence Modelling

JetWings

29

Page 30: Lecture # 05 Turbulence Modelling

19 May 2010

Resolved/Modeled/Total Reynolds stress (URANS)

Modeled

Resolved

Total

Lecture # 05 Turbulence Modelling

JetWings

30

Page 31: Lecture # 05 Turbulence Modelling

19 May 2010

Deep insight into RANS/DES (Point oscillation on free surface with power spectral analysis)

EFD DES URANS

2HZ

-5/3

Lecture # 05 Turbulence Modelling

JetWings

31

Page 32: Lecture # 05 Turbulence Modelling

19 May 2010

Direct Numerical Simulation (DNS)• DNS isto solve the Navier-Stokes equation directly without

averaging or approximation other than numerical discretizations whose errors can be estimated (V&V) and controlled.

• The domainof DNS must be at least as large as the physical domain or the largest turbulent eddy (scale L)

• The size of the gridmust be no larger than a viscously determined scale, Kolmogoroff scale, η

• The number of grid pointsin each direction must be at least L/ η

• The computational cost is proportional to

• Provide detailed information on flow field• Due to the computational cost, DNS is more likely to be a

research tool, not a design tool.

( ) 434/3 Re01.0Re ≈L

Lecture # 05 Turbulence Modelling

JetWings

32

Page 33: Lecture # 05 Turbulence Modelling

19 May 2010

Examples (Diffuser)

• Mean velocity predicted by V2f agreed very well with EFD data, particular the separation region is captured.

• K-ε model fails to predict the separation caused by adverse pressure gradient.

Lecture # 05 Turbulence Modelling

JetWings

33

Page 34: Lecture # 05 Turbulence Modelling

19 May 2010

Examples (Diffuser)

• TKE predicted by V2f agreed better with EFD data than k-ε model, particular the asymmetric distribution.

• Right column is for the skin friction coefficient on the lower wall, from which the separation and reattachment point can be found.

v2f

k-ε

2x/H

Lecture # 05 Turbulence Modelling

JetWings

34

Page 35: Lecture # 05 Turbulence Modelling

19 May 2010

References

1. J. H. Ferziger, M. Peric, “Computational Methods for FluidDynamics,” 3rd edition, Springer, 2002.

2. D. C. Wilcox, “Turbulence Modeling for CFD”, 1998.3. S.B.Pope, “Turbulent Flows,” Cambridge, 20004. P. A. Durbin and B. A. Pettersson Reif, “Statistical Theory and

Modeling for Turbulent Flows,” John Wiley & Sons, LTD, 2001.5. P.A. Davidson, “Turbulence: An Introduction for Scientists

and Engineers,” Oxford, 2004.

Lecture # 05 Turbulence Modelling

JetWings

35